• Nov 5th 2010, 05:42 PM
Sorena
ABC is a right triangle with ab (perpendicular) bc, AB = 50 cm, BC = 120 cm. Find the area of the largest rectangle that can be inscribed in (triangle) ABC with one of its corners at B.

Help pls.

• Nov 5th 2010, 06:02 PM
Educated
Code:

                c                    |\                 | \                  |  \                      |  \                120|____\                  | x  |\                 |    | \                 |  h|  \                 |____|___\                    b    50 ^  a                         |                         |                     (50-x)
Lets call the height if the rectangle h and the width of the rectangle x.

Using similar triangles, the height and width can be expressed as:

120 / 50 = h / (50-x)

Therefore $h = \frac{-12x}{5} + 120$

Now the area of a reactangle is width times height.

The width is x. The height is $\frac{-12x}{5} + 120$

Therefore, the area of the rectangle is $x(\frac{-12x}{5} + 120)$

The maximum area is found by finding the maximum turning point.

I'll leave that part for you.
• Nov 5th 2010, 06:05 PM
skeeter
the area function is a quadratic ... a derivative is not necessary to find the maximum.
• Nov 5th 2010, 06:11 PM
Educated
Quote:

Originally Posted by skeeter
the area function is a quadratic ... a derivative is not necessary to find the maximum.

I'm sorry, what do you mean?

2. Maximisation usually involves the gradient equalling 0, when the rate of change between both variables is at a minimum.
• Nov 5th 2010, 06:27 PM
Sorena
the final answer is 1500 cm^2. how?
• Nov 5th 2010, 06:32 PM
Educated
Do you know what differentiation is?
• Nov 5th 2010, 06:34 PM
Sorena
no
• Nov 5th 2010, 06:35 PM
Sorena
the thing is I have alot of questions right now, and i have to send them before midnight to get the mark, so im posting the ones that seem hard on first look on the internet and doing the other ones as fast as I can and then copying these ones. So I would appreciate it if you could just give me the answer please!!!

sry for the grammar !!!

no time!!!

• Nov 5th 2010, 06:38 PM
Educated
Quote:

Originally Posted by Sorena
the thing is I have alot of questions right now, and i have to send them before midnight to get the mark, so im posting the ones that seem hard on first look on the internet and doing the other ones as fast as I can and then copying these ones. So I would appreciate it if you could just give me the answer please!!!

sry for the grammar !!!

no time!!!

Forum rules.

You cannot get other people to do work for you that count for a mark/grade. (Rule 6)
• Nov 5th 2010, 06:44 PM
Sorena
ok just tell me how can i solve this using a quadratic equation?
• Nov 5th 2010, 06:45 PM
Educated
Maximisation and minimisation cannot be solved using a quadratic. Calculus would be involved.

And no, I will not give you the answer. You have to work that out yourself.
• Nov 5th 2010, 07:53 PM
mr fantastic
Quote:

Originally Posted by Educated
Maximisation and minimisation cannot be solved using a quadratic. Calculus would be involved.

And no, I will not give you the answer. You have to work that out yourself.

Calculus is not required. Post #3 makes that clear.

In support of the sentence in red: Teachers expect questions that form part of an assessment that contributes towards the final grade of a student to be the work of that student and not the work of others. For that reason, MHF policy is to not knowingly help with such questions.

Quote:

Originally Posted by Educated
I'm sorry, what do you mean?

2. Maximisation usually involves the gradient equalling 0, when the rate of change between both variables is at a minimum.

The turning point of a quadratic function can be found without using calculus. The OP ought to review his/her notes on the technique.